Key Arrow Patterns in Drawing Resonance Structures (Vid 2/4) 

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Check out NPTEL's videos, they have quite a bunch of videos about organic chemistry, even some more advanced ones, and they're all from India ;)

@@bonbonpony But Leah is the best. She is LITERALLY the best!!

@@alirezasadeghifar3815 I'm not saying that she isn't ;) I wouldn't personally say "best" (maybe "best on RU-vid), but yeah, she's pretty much up there… If only she didn't use that canned reply to each question about tutoring (even if people don't ask for any tutoring)… :q I would also recommend "freelanceteach" channel on RU-vid, that dude is also a very good teacher, with tons of good videos on his channel that are all freely available on a "pay if you like" basis.

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No, the left-most carbon only has 2 hydrogen atoms bound to it. The pi bond resonated away from that carbon, and no new bonds to hydrogen came in to replace it. Since there are only three total bonds formed by that carbon, we have a formal charge of +1.

That's correct. By convention, in bond-line notation, we do not show a hydrogen atom bonded to a carbon atom. It is most definitely there, even though not explicitly drawn.

Not a Dr, just Leah :) You can find the entire series along with the Resonance Guide and Practice Quiz on my website: leah4sci.com/resonance

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Alas you have to memorize them. Typically atoms that are very small will have

First, could you tell me which specific example in the video you're referring to? I believe the answer to your question is going to lie in remembering to follow the octet rule. If referring to the structure around 9:30, then pushing the pi bond up towards the oxygen would violate the octet rule. Oxygen cannot be surrounded by more than 8 electrons.

I think that the "opposite directions" might be the key here, because my intuition tells me that you might be trying to move electrons "uphill" the electric potential difference. Electrons are negative, so they want to move towards positive charges (or partially positive), not away from it. Opposite charges attract, so they would rather move towards each other than away. Moving them apart would require energy input. (It's possible, but unlikely to occur "on its own" without any external influences.) It is this natural attraction of opposite charges that usually drives chemical reactions, so always look out for charges to guide you.

always! Charge cannot be created nor destroyed, only transferred

@@Leah4sci Thanks.

@@Leah4sci Technically, the charge is destroyed (uncharged) when positive charges cancel with negative charges ;> (inb4: yes, I know that electrons and protons are still there. But the charges are no more, unless you separate them again.)

Atoms that are highly electronegative don't want to be electron deficient. They are always going to choose to have a complete octet. While carbon can hold a positive charge with only three bonds, nitrogen and oxygen will still have a complete octet when positively charged. For example, nitrogen with four bonds and no lone pairs (i.e. ammonium) has a complete octet and is positively charged.

@@Leah4sci thank you so much you help me a lot 💚

Thanks for your question! The starting structure in key arrow pattern 4 is positively charged and has no lone pairs. Similarly to lone pairs, the negative pi electrons have a strong pull towards that positive charge and can move towards it. Forming a lone pair on the adjacent carbon and creating additional separation of charge is undesirable.

Why don't you understand?

@@Leah4sci just because loool --> "ight imma head out"

For help with this, I recommend joining the organic chemistry study hall ;) Full details: leah4sci.com/join

Yes. Start with looking for lone pairs, because there are some that have not been drawn. Then think on which of these lone pairs can move to form π bonds. Only one of the two atoms can move its lone pair, the other one is stuck. And if you look closely, there is a nice patch for the charges to flow from one end of the molecule to the other, made by conjugated double bonds ;) (interleaved with single bonds).

@@bonbonpony hey, I'm a year late,but would you mind sharing how you solved the problem given at the end? I tried too,but since there's no solution available here to match, I'm not quite sure. Thank you.

@@bonbonpony oxygen,right? Nitrogen is stuck because carbon will not kick anything out?

@@thecosmos7671 Oxygen already has a π bond with carbon, so it can't use any of its lone pairs for resonance (also they are not in the plane of other π orbitals so they can't interact with them). If anything, the oxygen could only possibly withdraw its π bond to get a third lone pair and a negative charge, but it won't do that just like that - something would have to push those electrons towards it. Nitrogen, on the other hoof, has a lone pair which is in one of its sp³-hybridized orbitals, so it can donate this lone pair to some other atom if there's a π orbital nearby (and there is: a double bond between the two carbons next to it). There's actually a nice conjugated system of those π orbitals that goes from the nitrogen to the oxygen like a wire, single and double bond interchanged. And this is where the electrons will flow through, pushing each other like dominoes, starting from the nitrogen's lone pair, and ending up on the oxygen (which is quite electronegative and just loves to carry negative charges ;) )

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The lone pair probably came from the breaking of some other hydrogen bond before hand and that bond became a lone pair on the carbon giving us that negative charge you se

Remember that hydrogen atoms are not explicitly drawn in bond-line structures. The presence of that lone pair on the starting molecule tells us that there is only one hydrogen atom bound to that particular carbon in addition to the lone pair of electrons.

@@zaminahmed297 I wonder what reaction might have led to such configuration. If we had an oxygen instead of that nitrogen, that would be a carbonyl, and the carbon with that extra lone pair would be its α carbon (the one right next to the carbonyl carbon), and hydrogens on the α carbon is slightly acidic. It can in fact be knocked off in a presence of a strong base. Could the same reasoning still be applied if we have a nitrogen there instead? (that is, an enamine instead of a carbonyl)

The easiest way to see it, is to just follow the arrows, because they show the motion of electrons. If an arrow ends on an atom that was previously neutral, it will make it negatively charged, because the arrow brought one extra electron to it. If the atom started as positive, the arrow would make it neutral (because the negative electron that comes in will cancel the positive charge). No need to calculate formal charges all the time if you can just follow the arrows ;) But if you didn't have the arrows, just the end result, then yeah, you would have to count the electrons around this carbon yourself. Drawing the hidden links to hydrogens might help. In this case, there's only one hidden hydrogen. Why? Well, if the carbon is happy with this end result, it must have a full octet. We have two σ bonds to other carbons, and one lone pair, so there's only 2 more electrons missing for it to have a full octet - that's the σ bond to the hidden hydrogen. Now count how many extra electrons the carbon has: normally it would have 4, but here it has 5 (one in the σ bond on the left, one in the σ bond on the right, one in the σ bond with the hidden hydrogen, and the two from the lone pair), therefore it must have a negative charge.

Almost! It has 4 valence electrons but is bound to two carbons, TWO hydrogens, and has a lone pair. To learn more about formal charge and see a shortcut for calculating it, go to leah4sci.com/formal-charge-formula-and-shortcut/

@@Leah4sci Thanks so much for replying 🙂 Your videos are great.

Carbon #4 has 3 Hydrogen attached to it therefore it's sp3 Hope that helps!

sp3 carbons have 4 groups around them, bonds or lone pairs. In this case the one with the negative electrons has 2 bonds to carbon, one pair in the p-orbital, and an invisible hydrogen

Because when the electrons from the lone pair on the oxygen move down towards the carbon to make a double bond with it, at the same time they push away the electrons that the carbon already has in its preëxisting π bond with another carbon, basically breaking this π bond. (π bonds can be broken rather easily, they're more mobile; σ bonds not so easily) If that other π bond wasn't there, this trick couldn't be done (the carbon wouldn't be sp²-hybridized, but sp³-hybridized instead). So yeah, long story short, one π bond forms, another π bond breaks, and the carbon in the middle still has the same number of electrons afterwards. No octet rule broken here.

If you make bond to carbon AND have something else to kick out, you're simply swapping pi bonds rather than creating a 5th bond to the atom

I know youre a year out now... but that carbon had a hydrogen on it, you cant double bond that.

Carbon must follow the octet rule. It can only have a total of eight electrons in its surrounding orbitals. In the starting molecule at 2:28, the carbon in question has 2 bonds to other carbons, 1 bond to oxygen, and 1 bond to hydrogen. That already adds to eight electrons. Resonating electrons into a pi bond with oxygen would be impossible and would cause carbon to exceed the octet rule.

The number 12 doesn't come from anywhere in particular. Just know that anything larger than that is too large to sustain. We want to be as close to an octet as possible, but within reason. For your second time stamp, I don't see any triple bond, so I'm unsure of what second pi bond you're speaking of. Keep in mind that I don't offer tutoring over social media. For help with questions like this and more, I recommend joining the organic chemistry study hall. Details: leah4sci.com/join or contact me through my website leah4sci.com/contact/

@@Leah4sci I don't request tutoring, I don't need one. I'm just asking a simple question, something that usually can be answered in less than a minute. And I'm rather unfond of people who doesn't lift a finger until you paid them first :q because they don't have monopoly on knowledge, and if they don't want to share it, someone else will. (Let alone that they also forget how _they_ got their knowledge to begin with - from someone who shared it with them in the past. So it's like taking but not giving back.) As for the π bond I was asking about: I meant a resonance that switches between a triple bond and a double bond (i.e.: two π bonds vs. one π bond).

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Thank you!